/sci/ - Probability

Name: Anonymous 2011-05-29 20:03

I'm trying to teach myself probability, but it's really difficult for me to understand.

I'm trying this question from a textbook I've downloaded online, and I just can't seem to wrap my head around it.

Question 1. a) There is a family of 3 children. You saw one of the children, and it's a girl. What's the probability that all of them are girls?

1. b) You saw another child. You're not sure if it's the same girl you saw. What's the probability that all of the children are girls? Is it the same as question 1. a)?

Name: Anonymous 2013-05-30 16:22

P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(B|A)P(A))}{P(B)}

1.a)

P(3G|1G) = \frac{P(1G|3G).P(3G)}{P(1G)}

P(3G|1G) = \frac{1 \left (\frac{1}{2} \right )^{3}}{3 \left (\frac{1}{2} \right )^{3}}

P(3G|1G) = \frac{1}{3}

1.b)

\frac{P(3G|1G)}{3} + \frac{2P(3G|2G)}{3} = \frac{\frac{1}{3}}{3} + \frac{2}{3}\frac{P(2G|3G).P(3G)}{P(2G)}

= \frac{1}{9} + \frac{2 \left (\frac{1}{2} \right )^{3}}{9 \left (\frac{1}{2} \right )^{2}}

= \frac{1}{9} + \frac{1}{9}

= \frac{2}{9}

Probability

1 Name: Anonymous 2011-05-29 20:03

19 Name: Anonymous 2013-05-30 16:22

Name: Anonymous 2011-05-29 20:03

Name: Anonymous 2013-05-30 16:22